Sequences & Series
Alternating Series — e Expansion
nta_pyq_2026_jan
Grade 11
Question:
The value of $\displaystyle\sum_{k=1}^\infty(-1)^{k+1}\left(\dfrac{k(k+1)}{k!}\right)$ is
$2/e$
$1/e$
$e/2$
$\sqrt{e}$
Step-by-Step Solution
Key Concept: $\tfrac{k(k+1)}{k!}=\tfrac{k+1}{(k-1)!}$. Let $j=k-1$: sum $=\sum_{j=0}^\infty(-1)^j\tfrac{j+2}{j!}=\sum_{j=0}^\infty(-1)^j\tfrac{j}{j!}+2\sum_{j=0}^\infty\tfrac{(-1)^j}{j!}$.
$1/e$.
Correct Answer: 2